Impulse and Momentum

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln

Instructional Materials in Physics and Calculus-Based General Physics Astronomy

1975

IMPULSE AND MOMENTUM

Follow this and additional works at: https://digitalcommons.unl.edu/calculusbasedphysics

Part of the Other Physics Commons

"IMPULSE AND MOMENTUM" (1975). Calculus-Based General Physics. 18. https://digitalcommons.unl.edu/calculusbasedphysics/18

This Article is brought to you for free and open access by the Instructional Materials in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Calculus-Based General Physics by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Module -- 1 STUDY GUIDE

IMPULSE AND MOMENTUM

I NTRODUCTI ON You have already learned that you stub your toe harder trying to kick larger masses. Now imagine another unpleasant activity: catching a bowling ball. This gets harder to do as the ball is dropped from higher places. The diffi culty depends both on the ball's mass and its velocity just before you apply the stopping force. This force can be applied in different ways. Any winner of an egg-throwing contest will tell you the way to stop an object with the least force is to spread the stopping process out over a maximum time. Th is modu 1e wi 11 deve lop the above "folk phys i cs" into a sys tern of concepts and equations; and even a new law that is believed to be more fundamental than the laws from which it will be derived. This wonderful anomaly will not be further explored in this module, but it does indicate some of the philo sophical richness and curiosity that continues to be part of this science. The concepts are "center of mass" and "linear momentum"; the law is called "conservation of linear momentum."

PREREQU IS ITES

Before you begin this module, Location of you should bi able to: Prerequisite Content *Describe the motion of a body moving in a plane Planar Motion (needed for Objectives 1 to 3 of this module) Modul e *Solve problems requiring the application of Newton's Laws Newton's second and third laws (needed for Module Objectives 3 and 4 of this module)

LEARNING OBJECTIVES After you have mastered the content of this module, you will be able to: 1. Center of mass - Write the formulas for the center of mass (c.m.) of a system and explain all the terms. Write the formulas for the linear momentum of a system. Explain all the terms. 2. Linear momentum - Given the masses, positions, and velocities of all particles in a system, find the position and velocity of the center of mass, and the total (vector) linear momentum. STUDY GUIDE: Impulse and Momentum 2 3. Impulse - Given a force versus time graph or function for a system, calcu late the change of the system's linear momentum. 4. Linear-momentum conservation - Recognize conditions for which the linear momentum of a system is conserved. STUDY GUIDE: Impulse and Momentum 3(B 1)

TEXT: Frederick J. Bueche, Introduction to Physics for Scientists and Engineers (McGraw-Hill, New York, 1975), second edition

SUGGESTED STUDY PROCEDURE Read Chapter 9, Section 9.1 first. This explains the center of mass. Then read Chapter 7, Section 7.1, which introduces linear momentum and impulse and the relation between them. Now read Section 9.2, which partially shows how to calculate the velocity of a system's center of mass; You solve E~ (9.3) for vx(c.m.) and similar but unspecified equations for vy(c.m.) and vz(c.m.)' The v~locity comp~nents are A then combined into a single velocity: vc.m. = vx('c.m.)i + vy(c.m.)j + vz(c.m.)k. This section also relates the external forces acting on a system to the motion of the system's center of mass. Although Section 7.1 describes a single particle, Sections 9.1 and 9.2 have shown how to reduce a system of masses into a somewhat equivalent single particle. This particle has the total mass of the system and is located at the center of mass of the system. It is equivalent only in that it has the same linear momentum as the system. If you do not need to know details of the internal structure of the system, you can apply the ideas of Section 7.1 to systems of masses by treating them as a single particle located at the center of mass. For example, the earth is regarded as a particle by the scientists who calculate very high-altitude satellite orbits, but such an earth would be without meaning to most geologists. BUECHE Objective Problems with Assigned Problems Additional Number Readings Solutions Problems Study Text* Study Text* Guide (Ill us. ) Guide (Ill us.) Secs. 9. 1 , A A 7.1 2 Secs. 9. 1 , A, B 9.1 A, B 9.1 Chap. 9, 9.2 Probs. 1, 3, 4 3 Secs. 7. 1 , C 7. 1 , C 7. 1 , Chap. 7, 7.2 7.2 7.2 Quest.* 10, 12 4 Sec. 7.4 D 7.3, D 7.3, Chap. 7, 7.4, 7.4, Quest.* 5, 7 9.1 9.1 *Illus. = rllustration3s). Quest. = Question(s). STUDY GUIDE: Impulse and Momentum 3{B 2) Continue to read Sections 7.2, 7.3, and 7.4. Keep in mind that now when the text specifies a particle it can also be interpreted as referring to the c.m. of a system of masses. Read the General Comments; and read and understand how to solve the problem set. If you need additional help work some of the Additional Problems. Try the Practice Test. STUDY GUIDE: Impulse and Momentum 3(HR 1)

TEXT: David Halliday and Robert Resnick, Fundamentals of Physics (Wiley, New York, 1970; revised printing, 1974)

SUGGESTED STUDY PROCEDURE Read all of Chapter 8. Understand and know Eq. (8-3b). It contains all the previous equations in this chapter. Equation (8-4b) is Eq. (8-3b) in calculus form. The next important equation is (8-8). It also sums up the arguments of Section 8-2. Section 8-3 starts out with a definition you must memorize: the linear momentum of a particle, Eq. (8-9). Equation (8-10) will be used in Sections 8-4 and 8-5. It is a statement of Newton's second law. Section 8-4 shows how to calculate the total linear momentum ~f a system of particles: Eq. (8-12). Equation (8-13) can be used to find vc.m.. Equation (8-10) appears again in a more restricted form as Eq. (8-15): the internal forces have been eliminated. You should know why. Section 8-5 begins with the important equations describing the conservation of linear momentum, but paradoxically does not give them numbers. Read and understand Examples 1 to 6 in this chapter. Then read Section 9-2 in Chapter 9. Read the General Comments; and read and understand how to solve the Problem Set. If you need additional help, work some of the Additional Problems. Try the practice test.

HALL! DAY AND RESNI CK Objective Problems with Assigned Problems Additi ona 1 Number Readings Solutions Problems Study Text Guide Study Guide 1 Sec. 8-1 A

2 Secs. 8-1, A, B Chap. 8, A, B Chap. 8, 8-2, 8-4 Ex. * 1, Quest. 1 to 2, 3 4, 10; Probs. 1 to 13, 21, 22

3 Sec. 9-2 C Chap. 9, C Chap. 9, Ex. 1 to 4 Quest. 3,4, 8; Probs. 1 to 13 4 Sec. 8-5 D Chap. 8, o Chap. 8, Ex. 4, 5, Ques t. 5, 8, 6 9; Probs. 23 to 39 *Ex. = Example(s). Quest. = Question(s). STUDY GUIDE: Impulse and Momentum 3 (SZ 1)

TEXT: Francis Weston Sears and Mark W. Zemansky, University Physics (Addison Wesley, Reading, Mass., 1970), fourth edition

SUGGESTED STUDY PROCEDURE

f 'i University Physics does not include center of mass, and therefore other texts must be suggested for this important topic. Read the indicated sections in one of the three texts listed below. Author and Text Topic Section Frederick J. Bueche, Introduction to Ph~sics center of mass 9.1 for Scientists and Engineers {~cGraw- motion of c.m. 9.2 Hil', New York, 1975), second edition David Halliday and Robert Resnick, center of mass 8-1 Fundamentals of Ph~sics (Wiley, New York, motion of c.m. 8-2, 8-4 1970; revised printing, 1974) motion of c.m. 6-1 , 8-4 center of mass 6-2

SEARS AND ZEMANSKY Objective Problems with Assigned Number Readings Solutions Problems Additional Problems t Study Text Study Guide Guide Use one of the A Ex.* 1, A readi ngs recom 2 (Sec. mended above. 8-2) Sec. 8-1 2 Use one of the A, B A, B B: Chap. 9, 1, 3, 4 (non readings recom calculus), 5-8 (calculus) mended above HR: Chap. 8, Quest.* 1-3, 10; Probs. 1-13, Ex. 1-3 WS: 6-1, 6-3 to 6-8 (calculus)

3 Sec. 8-1 C Ex. 1 C 8-3, 8-4 (calculus) (Sec. 8-1) 4 Sec. 8-2 D D 8-5 *Ex. = Example(s). Quest. = Question(s). tB = Bueche. HR = Halliday and Resnick. WS = Weidner and Sells. STUDY GUIDE: Impulse and Momentum 3(SZ 2) The above readings develop the idea of replacing a system of masses by an imaginary but somewhat equivalent mass particle. This particle has the same total mass as the system and is located at a place called the center of mass (c.m.) of the system. In Chapter 8, Sections 8-1 and 8-2, University Physics develops the ideas of momentum and impulse for particles, and these concepts can be applied to all systems of masses since you now know how to reduce systems to equivalent particles. 1 Read Chapter 8. Examples 1 and 2 in Section 8-2 show that the momentum of a system of particles is the sum (vector) of the momenta of the particles. Read the General Comments; and read and understand how to solve the Problem Set. If you need additional help, work some of the Additional Problems. Try the Practice Test. STUDY GUIDE: Impulse and Momentum 3(WS 1)

TEXT: Richard T. Weidner and Robert L. Sells, Elementar Classical Physics (Allyn and Bacon, Boston, 1973), second edition, v01. 1

SUGGESTED STUDY PROCEDURE Read Chapter 5, Section 5-5. This section defines the linear momentum of a particle: Eq. (5-2). It furthermore shows by a worked example that in a aarticular collision the total linear momentum of the two colliding particles oes not change. The total linear momentum of a system of particles is defined in Eq. (5-5). Work through Examples 5-1 to 5-3. You should draw figures showing the linear-momentum vectors of the systems before and after the collision. Now read Chapter 6, Sections 6-1 and 6-2. These sections show you how to calculate the velocity and the location of a system's center of mass (c.m.). They also give you a method of reducing any collection of masses to a single somewhat equivalent particle located at the c.m. Now you have the tools to apply any rules for particles to collections of masses. The remainder of this module will show you how to use these tools. Read Section 7-4 in Chapter 7. Here you are shown the relationship between the linear momentum of a system and the forces acting on the system: Eq. (7-6). Example 7-3 should be understood. Read Chapter 8, Section 8-4. This is an important section because it tells you the general requirements for a system in order that its linear momentum be conserved. Example 8-8 is a good review of the ideas presented in this module. There are summaries at the ends of the chapters that gather the ideas into a few lines. Read the General Comments; and read and understand how to solve the Problem Set. If you need additional help, work some of the Additional Problems. Try the Practice Test. WEIDNER AND SELLS Objective Problems with Assigned Problems Additional Number Readi ngs Solutions Problems Study Text Study Text Guide Guide i " Sec. 5-5, Chap. A A 6, Sec. 8-4 2 Sec. 5-5, Chap. A, B A 5-3, 6-1 to 6, Sec. 8-4 6-8, 6-5 (calculus) 3 Sec. 7-4 C C 7-2 to 7-9, 7-11, 7-12 4 Sec. 8-4 D D STUDY GUIDE: Impulse and Momentum 4

GENERAL COMMENTS

1. Center of Mass The center of mass (c.m.) of a system is an intrinsic property of that system. Although the formulas for the position of the c.m. give it as the distance from the origin of a coordinate frame, the position of the c.m. does not move with respect to the system if the origin is moved. The distances from the origin to the c.m. will change, of course. But this is only because the origin or the system as a whole has moved. This property allows you to pick any point as the origin in a problem where you have to find a c.m. A clever choice may simplify your calculation. Look for a symmetry axis in your system and place the origin somewhere on it. Or you can place the origin on one of the system's particles.

2. Linear Momentum Linear-momentum vectors are added like any other vectors, but they exist in linear-momentum space (see Figure 1). Two-dimensional examples of a linear momentum space and an ordinary coordinate space are shown below. Points in ordinary space show position, and vectors in this space show displacements. Points and vectors in linear-momentum space represent linear momenta. You have no information about the position of anything in linear-momentum space. Sometimes when solving problems you may draw displacement and linear-momentum vectors in the same figure and not realize that they are superimposing ordinary and linear-momentum spaces. Watch out! This can lead to mistakes such as attempts to add displacement and linear-momentum vectors.

(x, y) (P,P) x y

P x x

Momentum Space Coordinate Space Figure 1 STUDY GUIDE: Impulse and Momentum 5

3. Coordinate System you will learn that the total linear momentum of a system is the product of the total mass of the system and the velocity of the system's c.m. If the system has zero resultant external forces acting on it, its c.m. is not accelerated. Thus the c.m. moves at constant velocity, and a coordinate system whose origin is placed at the c.m. and moves with it will be an iner tial coordinate system. Furthermore, the total linear momentum of the system relative to that origin is zero because the velocity of the c.m. relative to the origin (at the c.m.) is zero. Here is another case where a good choice of coordinate-system origin may sometimes simplify a problem. By placing the origin at the system's c.m. (providing the c.m. is not accelerating) you can use the fact that the total linear momentum of the system is zero. This placement of the coordinate system's origin establishes what is called the center-of-mass coordinate system or center-of-mass reference frame.

PROBLEM SET WITH SOLUTIONS A(l). (a) A system of several particles is shown in Figure 2. You are told the mass of the particles and their position coordinates relative to the coordinate axis. Explain how to find the center of mass of the system. (b) At the instant shown in Figure 2 the particles are in motion. m l is moving upward (+z direction) with speed v . m is moving to l 2 the left (-y direction) with speed v . m3 is moving toward m with 2 l speed v3. Explain how to find the total linear momentum of the system. z

-~ I X z zzl emz x Fi gure 2 STUDY GUIDE: Impulse and Momentum 6 Solution (a) By inspection of the figure you can see that the c.m. will be at a place that has x, y, and z coordinates. Your texts give formulas for each of these coordinates:

N N = 1 N x = L x.m./ L m. M L m.x. = X, c.m. i=l 1 1 i=l 1 i=l 1 1

N N N y = L y.m./ L m. = 1 L m.y. = Y, c.m. i=l 1 1 i=l 1 M 1=. 1 1 1

N N N z = L z.m./ L m. = 1 L m.z. = Z. c.m. i=l 1 1 i=l 1 M 1=. 1 1 1 Since this problem uses particles and not extended bodies you can use the c.m. formulas for particles. If the masses were not particles you would have to use the calculus formulas for c.m. coordinates. The c.m. of the system (see Figure 3) is located at the point having coordinates (xc.m.' Yc.m.' zc.m.)· You can also express the position of the c.m. by specifying the vector rc.m.' Use unit vectors and write

-+ A A A r c.m. = xC.m. i + y c.m. j + z c.m. k. The problem cuuld also have been given to you in terms of the masses and their position (or displacement) vectors. You would then have the additional first step of resolving the position vectors into their components, i.e., finding the xl' Yl' zl' x2'···· z rC.M.

I zcm ~----~~~~------yI / I / ___ ~/ xcm p Ycm Y Fi gure. 3 x Fi gure 4 STUDY GUIDE: Impulse and Momentum 7 (b) The total linear momentum of the system is the sum (vector) of the momenta of its parts. In this problem you must calculate the linear momentum of each particle and then add them to obtain the total linear momentum. The momentum of each particle is p = mv, where m is the mass and v is its velocity. The momenta vectors would look as shown in Figure 4, and the total linear momentum is

"* -+ -+ -+ -+ -+ -+ -+ ~ = Pl + P2 + P3 = MV x = ~lvlx + rn2v2x + m3v3x ' Since the three momentum vectors of the particles are not collinear nor even coplanar, the best way to add them is by their components. Resolve the velocity vectors into their x, y, and z components and add them:

Vx = vlx + v2x + v3x ' vY --v1y + v2y + v3y'

V = v + v + v ' MV z lz 2z 3z z Then

A A A + j + V = Vx i Vy Vz k, MV p Y y and fi na lly Fi gure 5 P = MV.

The total linear momentum and its components would look as shown in Fig~re 5. Although it's not asked for in this problem, you should recognize that V is the velocity of the center of mass of the system.

8(2). Three particles are moving radially outward from the coordinate origin, at angles of 120 0 to one another in the xy plane. Their masses are ml = 1.00 kg, m2 = 2.00 kg, and m3 = 0.80 kg, and their speeds are vl = 6.0 mis, v2 = 2.00 mis, and v3 = 10.0 mis, respectively. (a) Sketch the system in a coordinate frame. (b) Which particle has the momentum of greatest magnitude? (c) What is the total momentum of the three-particle system? (d) Find the velocity of the C.m. Solution (a) Draw an xy coordinate system and show the particles and their motions on it (Figure 6). There are many ways to place the particles on the coordinate system. In Figure 6 one of the partic1e ' s trajectories was placed along the y axis. This will save effort later if it becomes necessary to resolve the velocities or momenta into components in the x andy directions. STUDY GUIDE: Impulse and Momentum 8 Also draw the momentum vectors for the three particles (Figure 7). The Ma.gnitudes may be wrong in this figure because the linear momenta have not yet been calculated; but this gives you an idea of the momenta and their directions. Now start answering the questions. (b) Linear momentum is p = mt, and its magnitude is p = mv. For each particle:

P1 = ml v1 = (1.00 kg)(6.0 m/s) = 6.0 kg mis,

P2 = m2v2 = (2.00 kg)(2.00 m/s) = 4.0 kg mis,

P3 = m3 v3 = (0.80 kg)(10.0 m/s) = 8.0 kg m/s.

Particle 3 has the largest linear-momentum magnitude. We can now redraw Figure 7 to scale as Figure 8.

---+-.,.~.-.,t--- p x x

Figure 6 Figure 7

rs1

p x

Figure 8 STUDY GUIDE: Impulse and Momentum 9 (c) The total linear momentum of the system is "*t' = -+Pl + -+P2 + -+P3' In pictori~l form this addition is shown in Figure 9, and you see from inspec tion that P ~ O. Pity. Unless you prefer to work directly with the polygon in Figure 9 the best method to find P is to find its x and y components and add (vector) them. This will require you first to resolve the momenta of the three particles into their x and y components (refer to Figure 7): A A A j Pl = Plj = mlv l = (6.0j) kg mis,

A A 0 0 P2 = P2 cos (30 )i + P2 sin (30 )(-j)

A A = [(8.0)(O.866)i - (8.0)(O.500)j] kg mis,

A A P P cos (30 0 )(-i) + P sin (30 0 )(-j) 3 = 3 3 A A = [(4.0)(O.866)i - (4.0)(O.500)j] kg m/s. p - Figure 9 Use

Px = Pl x + P2x + P3x = 0 + (8.0)(0.866) - (4.0)(0.866)

A = 3.5i kg m/s. Similarly

Py = Ply + P2y + P3y = [6.0 - (8.0)(0.500) - (4.0)(O.500)i] kg mls =0 for this case. Thus for the total linear momentum, P = [(2.00)(l.732)i] = 3.5i kg m/s. (d) The velocity of the center of mass is the total linear momentum of the system divided by the system's mass:

-+ "* -+ -+ Vc.m. = t'/M= Emv/Em= V orMv(.x c.m. )=mlvl+ooo+mv.x n xn

Thus Vc.m. or V = [(2.00)(l.732)i kg m/s]/3.8 kg = (O.53)(l.732)i mls = 0.92i mis, Vx(c.m.) = [(2.00)(1.732) kg m/s]/3.8 kg = 0.92 mis, which is the magnitude of the velocity of the C.m. STUDY GUIDE: Impulse and Momentum 10

C(2). A croquet ball (mass 0.50 kg) initially at rest is struck by a mallet, receiving the impulse shown in Figure 10. What is the ball's velocity just after the force has become zero? Assume the graph is a parabola.

2.000 Figure 10

15"00 Figure 11 F(N) 1000 )

500 initi ally after force becomes zero

t(ms) Solution This is an impulse problem. The croquet ball's initial velocity (and there fore its initial momentum) is zero, and you want to find its final velocity. Since you are given the ball's mass, you must find its final momentum and divide by the mass: vf = Pf/m. Note that the problem is one dimensional, and you are not told the direction the ball rolls. However, you are asked for the ball's velocity (vector), and you must just assume a direction such as "horizontally to the right," and proceed to find the speed. Start with t f dt -r -r -r J F = ~p = mV - mv , ti f i

and apply this equation to the croquet ball. Since Vi = 0, you can solve for algebraically: vf t v = ~ f f F dt. f t., You know m and the integral is the area unde~ the force versus time graph. Here are two ways to integrate this area: (1) Square coynting: See Figure 12. Each small square has an area of (100 N) (0.200 x 10- 3 ) = 0.0200 N s. Now you simply count the number of squares under the graph and multiply this by the area of one square. A count accurate to 1% STUDY GU I DE: Impul se and Momentum " gives 210 squares. Thus the total area is t f f F dt = (210 squares)(0.0200 N s/square) = 4.2 N s. ti Since this is a one-dimensional problem the vector notation has been dropped:

1 t f 4.2 N s vf = m{. , F dt and vf = 0.50 kg = 8.4 mise (2) Calculus: An equation for a parabola is y = kx 2• However, this parabola passes through the origin, but your graph does not have F at zero when t is zero; thus the equation F = kt2 will not work for you. The equation 3 (F - 2200 N) = k(t - 2.00 x 10- s)2 will work, if you evaluate k. Pick a point such as F = 0 and t = 0.50 x 10-3, and plug these values into the above equation. This gives

k = -9.78 x 108 N/s2. Now 8 3 F = (9.78 x 10 N/s2)(t - 2.00 x 10- s)2 + 2200 N and t 3.4 x 10-3 s f 8 3 f F dt = f 3 [(-9.78 x 10 N/s2)(t - 2.00 x 10- s)2 + 2200 N]dt t., 0.50 x 10- s -3 = (-9.78 x 108 N/s2) J3.4 x 10_ s (t _ 2.00 x 10-3 s)2 dt 3 0.50 x 10 s 3 3.4 x 10- s + (2200 N) f -3 dt. 0.50 x lOs 3 A change of variable will simplify the first integral. Let t - 2.00 x 10- s :: T.

Fi gure 12 STUDY GUIDE: Impulse and Momentum 12 Then dt = dT, and changing the limits of integration gives us 3 3 8 2 1.4 x 10- 2 3.4 x 10- F dt = (-9.78 x 10 N s ) I -3 T dT + 2200 Nf -3 dt = 4.38 N s . . -1.50 x 10 0.50 x 10 Note here that this value is a bit more than the one obtained from square count ing. Eossibly the assumption that the graph was a parabola symmetric about 2 x 10 3 S was not correct. As before,

vf = 4.38 N s/0.50 kg = 8.76 m/s = 8.8 m/s.

0(2). A child runs and leaps into a stationary wagon. The wagon can roll without friction on the level, rough driveway, but it is not headed in the direction the child was running. The wagon and child move in the direction the wagon was pointed. (a) Is the linear momentum of the child-wagon system conserved in this process? Explain why. (b) Are any linear-momentum components of this system conserved? Explain why. Solution (a) Momentum is conserved if the total linear momentum of the system does not change. Before the child jumps on the wagon she alone has some linear momentum. The total linear momentum of the system then is in the direction the child is running. When the child is aboard the rolling wagon the total linear-momentum vector of the system points in the direction the wagon is rolling. This is not the direction in which the child was running, and thus the linear momentum of the system cannot be the same before and after the child jumped on the wagon. It is impossible for two vectors to be equal if they are not in the same direction. The linear momentum of the system has not been conserved. (b) Another way to identify momentum conservation is by the use of the equation Itf F dt = 6p. t. ext 1

I If Fext is zero, then so will be 6p. \ Furthermore, if there is any direction in which some component of t is Fex zero, the component of 6P in that direction will also be zero. In your problem the wagon rolls without friction in the direction it is pointed. There can be no external forces acting on the system in this direction, and the com ponent of the system's linear momentum in this direction ;s conserved. You should realize that there is a considerable external force on the system when the child lands ;n the wagon: the frictional force between the wheels and the STUDY GUIDE: Impulse and Momentum 13 rough driveway. This is the force that causes the momentum of the system to change. An often overlooked but not always trivial, direction in which momentum might be conserved is the vertical. The ~xternal v~rt1cal forces actin~ on the child-wagon system add to zero (assum1ng th~ Ch1ld s c:m. mo~es h?r1-. zontally); and the system's change in momentum 1n the ve~t1cal d1re~t1on 1S thus also zero. The system's linear-momentum component 1n the vert1cal direction remains zero during the process.

PRACTICE TEST

1. Explain how conservation of momentum applies to a handball bouncing off a wall. 2. Three particles floating in space are attached to one another with springs. Their masses are 5.0 kg, 7.3 kg, and 12.2 kg, respectively. One of the particles is hit by a meteorite. The force-time graph of this collision is shown in Figure 13. Calculate the change in momentum of the three particle system. 3. (a) Write the formula for the center of mass of any system. Explain what you would need to know about the system in order to calculate it. (b) Wrjte the formula for the linear momentum of any system whose parts are moving in straight lines and not rotating. Explain what you would need to know about the system in order to calculate it.

F(N

Figure 13 STUDY GUIDE: Impulse and Momentum 14

Practice Test Answers 1. The linear momentum of the ball is certainly not conserved. The ba11·s momentum has flipped direction owing to the force from the wall. If you include the wall and everything it is attached to in your system then momentum must be conserved: no external forces act. It is not easy to visualize the wa11·s momentum changing during the collision, but it does. Its great mass permits a very small velocity change.

t f 2. Use f F dt = ~P. t., ext The system is the three masses and the springs. The spring forces are inter nal. The only external force is provided by the meteorite. F dt = 6.0 x 10-3 Ns. Thus, ~p = 6.0 x 10-3 Ns in the direction of the external force.

+ " A A - . + . + k 3. ( a ) rc.m. - xc.m.' Yc.m.J zc.m.' where N N n x = ~ xm/ ~ m + ~ (f x dm/f dm), c.m. i=l i=l i=l N N n y = ~ ym/ ~ m + ~ (f y dm/f dm), c.m. i=l i=l i=l N N n z = ~ zm/ ~ m + ~ (f z dm/f dm) c.m. i=l i=l i=l for particles and extended bodies. You must know all the masses and positions of the particles; and the positions, shapes, and density distributions of the extended bodies. N -+ n (b) P = ~ mv + ~ mV i =1 ;=1 c.m. particles extended bodies

You must know the masses and velocities of the particles; and the masses and velocities (of any portion) of the extended bodies. IMPULSE AND MOMENTUM Date ------pass recycle Mastery Test Form A 1 2 3 4 Name ______Tutor ------1. (a) Write the formulas for the center of mass of any system. Explain all the terms.

(b) Write the formulas for the linear momentum of any system that h~s no rotation. Explain all the terms. 2. I

Two particles are moving apart as shown in the figure above. The mass ml is 5.0 kg and m2 is 3.00 kg; at the instant shown they are 6.0 m ?part. Each has a speed of 15.0 m/s. (a) Find the center of mass of this system at the instant shown. (b) Find the total linear momentum of this system at the instant shown. (c) Find the velocity of the center of mass of this system at the instant shown. 3. A 2.0-kg ball falls at a constant speed of 0.100 mls through a viscous fluid. What is the force of the fluid on the ball? 4. Attack or defend the following statement: If a system is made large enough its linear momentum is always conserved. Date ______IMPULSE AND MOMENTUM pass recycle Mastery Test Form B 1 2 3 4 Name ------Tutor ------1. (a) You are shown a system of particles and larger bodies in motion. What would you have to know about the system to calculate its center of mass? How would you calculate the center of mass of this system? (b) None of the bodies in the above system is rotating. What would you have to know about the system to calculate its linear momentum? How would you calculate the linear momentum of the system? 2. (a) Starting with a statement for the conservation of linear momentum, show that it takes an external force to accelerate the center of mass of a system with constant mass. (b) You are a prisoner in a 4.0-m-long boxcar whose frictionless wheels are 1.50 m from the top of a downhill grade (see Figure 1). If you can get the car to start rolling downhill, you can escape to friendly territory. The end of the car nearest the grade is stacked directly over the wheels with 1000 50-kg gold bars. The car has a mass of 40 000 kg. How many bars must you move to escape? Assume you can move the gold the full 4.0 m and ignore your mass. 3. A vertical rod is connected to a 40-kg particle (see Figure 2). The rod exerts a time-varying force on the particle, which can be calculated from the function

F = (200 + 300t)j N. (a) What is the force on the particle at 0 s? Do not neglect gravity. (b) Calculate the particle's change of linear momentum between the first and second seconds. (c) Is there a time when the linear momentum of the particle is not changing? If so, calculate this time. (d) Is there any component of the particle's linear momentum that is always conserved? If so, what is it and why is it always conserved?

y ~ particle

~ rod ------Mt------x Figure 1 Figure 2 IMPULSE AND MOMENTUM Date ------pass recycle Mastery Test Form C 1 2 3 4 Name Tutor ------1. (a) Write the formula for the center of mass of any system. Explain what you would need to know about the system in order to calculate it.

(b) Write the formula for the linear momentum of any system of nonrotating masses moving in straight lines. Explain what you would need to know about the system in order to calculate it. 2. A 30.0-kg particle is suspended by a string. The string is pulled upward and exerts a time-varying force on the particle. The magnitude of this force is given by the function F = (350 + 150t2) N. (a) What is the force on the particle at 0 s? (b) Calculate the change in the linear momentum of the particle between the second and third seconds. (c) Is the change of the particle's linear momentum per second constant? Why? 3. For which of the following systems (underlined) is linear momentum conserved? Justify your answers. (a) Two colliding billiard balls, rolling on a pool table. (b) A canoe with three nonpaddling occupants on a smoothly flowing river. IMPULSE AND MOMENTUM Date ------pass recycle l'lastery Test Form D 2 3 4

Name Tutor ------.------l. (a) Write the formulas for the center of mass of any system. Explain all the terms. / I (b) Write the formu 1as 1 i nea r momentum of any system that has no rotation. Explain terms.

2. ....lo. V, ~-t~

Two particles are moving apart as shown in the figure above. The mass m l is 5.0 kg and m2 is 3.00 kg; at the instant shown they are f.oc.a.J.l!c:.i at 31m O41J _A t ~ C. 171 1'"1..$6'(". ItlC./1o (a) Find the center of mass of this system at the instant shown. (b) Find the total linear momentum of this system at the instant shown. (c) Find the velocity of the center of mass of this system at the instant shown. 3. A man standing on a loading dock throws bags of sand into a passing truck. Discuss the conservation of momentum. Are all components of the linear momentum conserved? Calculate the change of momentum during the period 0-10 s for a 10 kg mass 4. .. acted upon by the force shown in the figure below.

t(S)

Courtesy of University of Missouri-Rolla IMPULSE AND MOMENTUM Date ------pass recycle Mastery Test Fonn E 2 3 4

Name ------Tutor ------1. (a) Write the fonnu1a for the center of mass of any system. Explain what f you would need to know about the system in order to calculate it.

F = (350 + SIII-¥)N (a) What is the force on the particle at 0 s? (b) Calculate the change in the linear momentum of the particle between the second and third seconds. (c) Is the change of the particle's linear momentum per second constant? Why?

3. A man stands on the back of a 2000 kg truck and throws 20 kg sandbags out the back with a velocity of 10 m/s' How many sandbags must he throw to get the truck velocity up to 1 m/s? Assume the truck wheels are frictionless. Neglect the change in the truck's mass due to the loss of sandbags.

4. What would be the effect of the man throwing sandbags off the side of the truck?

Courtesy of University of Missouri-Rolla IMPULSE AND MOMENTUM Date pass recycle

"1astery Test Form F 2 3 4 Name Tutor ______1. (a) You are shown a system of particles and larger bodies in motion. What would you have to know about the system to calculate its center of mass? HOI" would you calculate the center of mass of this system? (b) None of the bodies in the above system is rotating. What would you have to know about the system to calculate its linear momentum? How would you calculate the linear momentum of the system?

2. Assume you are standing (m=60 kg) standing at one end of a boxcar (m=6000 kg). If you move 5 m to the other end of the boxcar, how far will the boxcar move? Assume the wheels of the boxcar are frictionless.

3. A force t = (2001 + 100t.-~ N is exei'ted on a particle of mass 10 kg. Calculate the change in momentum of the body between the 5th and 10th seconds.

4. A man proposes to power his sail boat by mounting a fan on the top of the cabin and blowing air into the sail during still air periods. Draw a sketch and discuss the physics involved in this project.

Courtesy of University of Missouri-Rolla IMPULSE AND MOMENTUM A-l

MASTERY TEST GRADING KEY - Form A

What To Look For Solutions 1. System includes part l.(a) The answer must combine the c.m. coordinates icles and extended bodies. for all the masses into a sinqle set of + I( The student must show some coordinates: either X . .' Y . .' Zc.m. or r . . realization that the c m c m c m location of the c.m. is with respect to a set of N N n X = ~ Xlm./~ m. + ~ (fX dm/fdm) , axes. Two- or three c.m. i=l 'i=l 1 i=l dimensional answer. If in part (b) the student introduces the for Yc.m. and Zc.m. ,or troubles with nonrigid bodies, you must make a ~ A A A comment to him that this rc ..m = Xc.m. i + Yc.m. j + Zc.m. k. is too advanced for this course. + N + N (b) p = (.~ miv\i) particles + (.~ MVc m.) 1=1 J=l . extended bodies. Other forms and an answer in components are OK.

2. The student's choice 2.(a) y of coordinates. Comment on an awkward choice, and suggest a better one. However, a poor choice will not make the problem x wrong--. --Directions should be given for all vector quantities.

yc.m. = 0; placement of coordinate system.

xc.m. = (0 + m2x)/(m1 + m2) = (3.00 kg) (6.0 m)/8.0 kg = 2.3 m. I~PULSE AND MOMENTUM A-2

2. (b) P p. Y 1.

------~~~------Px j

p = P, + P2 = mlvl + m2v2· P = mlvlJ - m v2J = (m,v l - m2V2)J l 2 A = [(5.0 kg)(15.0 m/s) - (3.00 kg)(15.0 m/s)Jj

A = 30 j kg m/s.

3. The velocity of the ball is constant, and thus so is its linear momentum. The total external force on the ball is zero. There are two external forces acting on the ball: its weight and the force from the liquid. y f = W= mg = (2.00)(9.8)3 kg m/s 2 = 19.6j N.

------+~------x The statement is true. By increasing the size of the system, more and more of the w forces become internal forces. Some students are clever enough to attack the statement successfully. You must be equally clever in analyzing their arguments. For example, if it is argued that very small external forces will cause unmeasurable momentum changes to large masses, and therefore all the forces do not have to be internal forces, you must accept this. However, you might mention that an improvement in the technology of momentumometers might make their argument wrong. Do not accept arguments based on one particle universes. H~PULSE AND MOMENTUM 8-1

MASTERY TEST GRADING KEY - Form 8

What To Look For Solutions 1. System includes l.(a) The answer must combine the c.m. coordinates particles and extended for all the masses into a single set of -+ bodies. T~student must coordinates: either X . .' Y . .' Zc.m. or r . . show some realization that c m c m c m the location of the c.m. N N n is with respect to an axis. X = L X.m./ L m. + L (fX dm/fdm) , Two- or three-dimensional c.m. i=l 1 1 i=l 1 i=l answer. If in part (b) the student introduces the for Y and Z ,or troubles with nonrigid C.m. c.m. A A bodies, you must make a -+r =X i+Y '+Z k comment to him that this is c.m. c.m. c.m. J c.m.· too advanced for this course. You must know mass and position of all particles; and position, shape, and density distribution of the extended bodies.

N n (b) P = ( L m.V.) + ( L MV m) i=l 1 1 particles j=l c. 'extended bodies. You must know mass and velocity of all particles. For extended bodies you only need to know the total mass and the velocity of any part of the body.

2.(a) Start with Fext = M (dVc.m./dt) or

fFext dt = ~~ ~ ~(MVc.m.)' and argue that if the left-hand sides of these equations are not zero, neither will the right hand sides be zero. Recognize that the right hand sides, which all have a change in velocity, contain the acceleration of the center of mass. 2.(b) The student (b) The c.m. of the system does not move. As should realize that here the bars are moved to the left the car must roll is a case where the to the right. center of mass will not accelerate. Since it is X . . = (MCXC + MGXG)/(M + M) , initially at rest it will c m C G continue to be at rest ...... relative to some place X~c.m. = (MCXC + MGXG + M X )/(MC + MG + M ). outside the system. Of IMPULSE AND MOMENTUM 8-2

What To Look For Solutions course, once the front wheels move over the edge of the incline a net external force acts, I ( and the center of mass I accelerates. CM• S

k a

IE+-- x" I C x" ..I ___ ~ k I I

Neglect prisoner's mass.

~ X = X , c.m. c.m. xc = a/2 + b,

~ Xe = a/2,

~ ~ I X = a, Me + MG = Me + MG + M , '"

~ Me (a/2 +b) + MGb = Me (a/2) + 0 + Ma.

~ Solving for M :

~ M = (Meb + MGb)/a = 33 750 kg. IMPULSE AND MOMENTUM 8-3

What To Look For Solutions Number of blocks moved is n - 31 250 kg - 675 blocks. -50 kg/block -

3. Magnitudes and 3.(a) Two forces: y directions of vectors. Recognition that this is F mainly an impulse problem. In part (c) x there will be an instant when the particle1s momentum change is zero. w It is poor practice to say that momentum is ~ A ~ A conserved then. At t = 0, r = 200j N, W = mg(-j).

~ ~ A Total force = r + W= -192j N. t (b) Use t.f, f F dt = IIp,

2 s A A I [(-192 + 300t)j] dt = 258j kg m/s. 1 s

(d) There are no horizontal forces acting on the particle. Thus the horizontal components of the linear momentum remain constant. IF dt = llP and if F = 0, llP = o. IMPULSE AND MOMENTUM C-l

MASTERY TEST GRADING KEY - Form C

What To Look For Solutions 1. System includes l.(a) The answer must combine the c.m. particles and extended coordinates for all the masses into a single bodies. The student must set of coordinates: Either X ,Y ,Z show some realization that or -+r c.m. c.m. c.m. the location of the c.m. c.m. is with respect to an axis. N N n Two- or three-dimensional X = E X.m.1 E m. + E (IX dmlldm) answer. If in part (b) c.m. i=l 1 1 i=l 1 i=l the student introduces the troubles with nonrigid for Yc.m. and Zc.m. ,or bodies, you must make a + A '" '" comment to him that this r = X i + Y j + Z k. is too advanced for this c.m. c.m. c.m. c.m. course. You must know mass and position of all particles; and position, shape, and density distribution of the extended bodies. N n (b) ~ = ( E m.f.) + ( E Mf ) i=l 1 1particles j=l c.m'extended bodies You must know mass and velocity of all particles. For extended bodies you need only know the total mass and the velocity of any part of the body.

2. Magnitudes and 2.(a) Two forces: y directions of vectors. Recognition that this F is mainly an impulse problem. x

At t = 0, F = 350] N, W = mg(j).

or ~ A Force: ~ + W= 56j N.

t (b) Use f fF dt = ~~. t. 1

3 s 2 A A f [(56 + 150t )j] dt = [(56 + 950)]j A 2 s = l006j kg m/s. IMPULSE AND MOMENTUM C-2

What To Look For Solutions 3. The student must 3. Depending on the viewpoint, both yes and no understand that his answer can be justified: will depend on the existence (a) Yes - internal forces between tAe balls are of external forces acting involved and dominate at the moment of on the system. collision; external forces are comparatively small. No - over a slightly longer time interval, the interaction of the balls with the table must be considered; since the collision will result in some slippage of the balls on the table, they are subject to a friction force that may change the momentum of the two-ball system. (b) Yes - the net force is zero, vertical forces-of water and gravity cancel, no paddling means zero horizontal force; motion of river is smooth, does not give rise to force. No - if river flows around curve.